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Difference between revisions of "1989 AHSME Problems/Problem 2"

(Created page with "<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=</math> (A) <math>\frac{1}{5}</math>\hskip.5in (B)<math>\frac{1}{4}</math>\hskip.5in (C)<math>\frac{2}{7}</math>\hskip.5in (D)<math>\...")
 
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<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=</math>
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<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}</math> and hence the answer is D.
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(A) <math>\frac{1}{5}</math>\hskip.5in (B)<math>\frac{1}{4}</math>\hskip.5in    (C)<math>\frac{2}{7}</math>\hskip.5in    (D)<math>\frac{5}{12}</math> \hskip.5in  (E)<math>\frac{7}{12}</math>
 

Revision as of 09:20, 19 February 2012

$\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}$ and hence the answer is D.

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